Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Physics
Chapter
Oscillations

JEE Advanced 2025 · Paper 1 · Physics · Oscillations

The center of a disk of radius r and mass m is attached to a spring...

The center of a disk of radius rr and mass mm is attached to a spring of spring constant kk, inside a ring of radius R>rR>r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as T=2πωT=\frac{2 \pi}{\omega}. The correct expression for ω\omega is ( gg is the acceleration due to gravity): The center of a disk of radius r and mass m is attached to a spring of spring constant k , inside a ring of radius R>r

Options

  1. A\sqrt{\frac{2}{3}\left(\frac{g}{R-r}+\frac{k}{m}\right)}
  2. B2g3(R−r)+km\sqrt{\frac{2 g}{3(R-r)}+\frac{k}{m}}
  3. C\sqrt{\frac{1}{6}\left(\frac{g}{R-r}+\frac{k}{m}\right)}
  4. D\sqrt{\frac{1}{4}\left(\frac{g}{R-r}+\frac{k}{m}\right)}

Correct answer

A. \sqrt{\frac{2}{3}\left(\frac{g}{R-r}+\frac{k}{m}\right)}

Step-by-step solution

The center of a disk of radius r and mass m is attached to a spring of spring constant k , inside a ring of radius R>r
E=12k(R−r)2θ2+mg(R−r)(1−cos⁡θ)+12mv2+12mr22ω2E=\frac{1}{2} k(R-r)^2 \theta^2+m g(R-r)(1-\cos \theta)+\frac{1}{2} m v^2+\frac{1}{2} \frac{m r^2}{2} \omega^2 Differentiating wrt t , 0=12k(R−r)2⋅2θ dθdt+mg(R−r)⋅ddt(2θ24)+12 m⋅2vdvdt+mr24⋅2ω dωdt⇒0=k(R−r)2θ dθdt+mg(R−r)θdθdt+mvdvdt+mr22ω dωdt\begin{aligned} 0 & =\frac{1}{2} \mathrm{k}(\mathrm{R}-\mathrm{r})^2 \cdot 2 \theta \frac{\mathrm{~d} \theta}{\mathrm{dt}}+\mathrm{mg}(\mathrm{R}-\mathrm{r}) \cdot \frac{\mathrm{d}}{\mathrm{dt}}\left(2 \frac{\theta^2}{4}\right)+\frac{1}{2} \mathrm{~m} \cdot 2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{mr}^2}{4} \cdot 2 \omega \frac{\mathrm{~d} \omega}{\mathrm{dt}} \\ \Rightarrow & 0=\mathrm{k}(\mathrm{R}-\mathrm{r})^2 \theta \frac{\mathrm{~d} \theta}{\mathrm{dt}}+\mathrm{mg}(\mathrm{R}-\mathrm{r}) \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}+\mathrm{mv} \frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{mr}^2}{2} \omega \frac{\mathrm{~d} \omega}{\mathrm{dt}}\end{aligned} Also, dθdt=V(R−r)⇒d2θdt2=1(R−r)dvdt=1R−ra\frac{d \theta}{d t}=\frac{V}{(R-r)} \Rightarrow \frac{d^2 \theta}{d t^2}=\frac{1}{(R-r)} \frac{d v}{d t}=\frac{1}{R-r} a∴k(R−r)2⋅θ VR−r+mg(R−r)θVR−r=−mvαr−mr22 Vrα⇒k(R−r)+mgθ=−32mrα⇒−[k(R−r)+mg]θ=32 m(R−r)d2θdt2⇒−23[km+gR−r]=d2θdt2\begin{aligned} & \therefore \mathrm{k}(\mathrm{R}-\mathrm{r})^2 \cdot \theta \frac{\mathrm{~V}}{\mathrm{R}-\mathrm{r}}+\mathrm{mg}(\mathrm{R}-\mathrm{r}) \theta \frac{\mathrm{V}}{\mathrm{R}-\mathrm{r}}=-\mathrm{mv} \alpha \mathrm{r}-\frac{\mathrm{mr}^2}{2} \frac{\mathrm{~V}}{\mathrm{r}} \alpha \\ & \Rightarrow \mathrm{k}(\mathrm{R}-\mathrm{r})+\mathrm{mg} \theta=-\frac{3}{2} \mathrm{mr} \alpha \\ & \Rightarrow-[\mathrm{k}(\mathrm{R}-\mathrm{r})+\mathrm{mg}] \theta=\frac{3}{2} \mathrm{~m}(\mathrm{R}-\mathrm{r}) \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2} \\ & \Rightarrow-\frac{2}{3}\left[\frac{\mathrm{k}}{\mathrm{m}}+\frac{\mathrm{g}}{\mathrm{R}-\mathrm{r}}\right]=\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\end{aligned}Compering with standard equation of SHM $\omega=\sqrt{\frac{2}{3}\left[\frac{\mathrm{k}}{\mathrm{m}}+\frac{\mathrm{g}}{\mathrm{R}-\mathrm{r}}\right]}$ Hence answer is option(A) $\mathrm{OR}$<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/W7rz1tmNqSgEYtx95pEk2h7vBmnz9dbCeSTzR6BhXDE.original.fullsize.png"><br>$\mathrm{kx}+\mathrm{mg} \sin \theta-$\mathrm{f}=\mathrm{ma}$\begin{aligned} & \Rightarrow \mathrm{kx}+\mathrm{mg} \frac{\mathrm{x}}{(\mathrm{R}-\mathrm{r})}-\mathrm{f}=\mathrm{ma} \\ & \mathrm{fr}=\frac{\mathrm{mr}^2}{2} \cdot \alpha \Rightarrow \mathrm{f}=\frac{\mathrm{ma}}{2} \\ & \therefore\left(\alpha+\frac{\mathrm{mg}}{\mathrm{R}-\mathrm{r}}\right) \mathrm{x}=\frac{3 \mathrm{ma}}{2} \\ & \therefore \omega=\sqrt{\frac{2}{3}\left[\frac{\mathrm{k}}{\mathrm{m}}+\frac{\mathrm{g}}{\mathrm{R}-\mathrm{r}}\right]}\end{aligned}$$
Practice Oscillations on Quantrex Academy →

More from Oscillations

A tank contains two immiscible liquids of densities 6ρ and 2ρ . The higher density liquid… JEE Advanced 2026The frequency of oscillation of a mass m suspended by a spring is v_1 . If the length of… JEE Main 2026A spring stretches by 2 mm when it is loaded with a mass of 200 g. From equilibrium… JEE Main 2026A particle is executing simple harmonic motion. Its amplitude is A and time period is 5… JEE Main 2026Match List-I with List-II. List-I List-II A. ^2 ω t I. Periodic with time period T =… JEE Main 2026A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in… JEE Main 2026The velocity of a particle executing simple harmonic motion along x -axis is described as… JEE Main 2026The equation of motion of a particle is given by x = a (50t + (π)/(3)) cm. The particle… JEE Main 2026 Full Oscillations list All JEE Advanced PYQs